A268336 -id:A268336 - cp's OEIS frontend

A267817 Numbers m that are divisible by A268336(m).

1, 2, 4, 6, 8, 10, 12, 16, 18, 20, 24, 30, 32, 36, 40, 42, 48, 50, 54, 60, 64, 68, 72, 78, 80, 84, 90, 96, 100, 108, 110, 114, 120, 126, 128, 136, 144, 150, 156, 160, 162, 168, 180, 192, 200, 204, 210, 216, 220, 222, 228, 234, 240

Offset: 1

Juri-Stepan Gerasimov, Feb 13 2016

nonn

Squarefree terms: 1, 2, 6, 10, 30, 42, 78, 110, 114, 210, 222, ...

			10 is in this sequence because 10/A268336(10) = 10/2 = 5.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Juri-Stepan Gerasimov, divisors (k^k mod n) of n, SeqFan list, Feb, 13, 2016

Cf. A174824, A268336.

is(n)=my(f=factor(n)[, 1],m=n); for(k=1, #f, m=lcm(m, f[k]-1)); m/=n; m && n%m==0 \\ Charles R Greathouse IV, Feb 22 2016

a(16) inserted by Charles R Greathouse IV, Feb 22 2016

A329338 a(n) = {1{0}^(A268336(n)-1)}^(n-1){1}{0}^A051903(n): upper bound for A329126(n).

Original entry on oeis.org

1, 110, 101010, 111100, 100010001000100010, 1111110, 10000010000010000010000010000010000010, 11111111000, 1010101010101010100, 10101010101010101010, 100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000100000000010

Offset: 1

M. F. Hasler, Nov 13 2019

nonn

This is the upper bound for A329126 as explained in the "FORMULA" there.

It is sharp for all n except 10, 14, 15, ...

Cf. A329126, A329000, A329339 (this converted from binary to decimal), A268336, A051903.

apply( {A329338(n)=my(k=lcm(lcm([p-1|p<-factor(n)[,1]]), n)/n); fromdigits(concat(vector(n, i, Vec(1, if(i1, vecmax(factor(n)[,2])+1)))))}, [1..16])

a(n) = A007088(A329339(n)), where A007088 = binary numbers and A329339(n) = 2^A051903(n)*(m^n-1)/(m-1) with m = 2^A268336(n).

A329339 a(n) = 2^A051903(n)Sum_{k=0..n-1} 2^(A268336(n)k): upper bound for A329000(n).

Original entry on oeis.org

1, 6, 42, 60, 139810, 126, 139620524162, 2040, 349524, 699050, 2537779500750160131246576896002, 16380, 44612382091907903486070965589630128805126146, 1256584717458, 153722867280912930, 1048560, 231587712222682663714935471840371426842813815977643091627066215779128553111554, 1048572

Offset: 1

M. F. Hasler, Nov 13 2019

nonn

This corresponds to the upper bound for A329000 as explained in the "FORMULA" for A329126.

Differs from A329000(n) for n = 10, 14, 15, ...

Cf. A329000, A329126, A329338 (a(n) written in binary), A067029, A051903.

apply( A329339(n)={my(m=2^(lcm(lcm(znstar(n)[2]),n)/n)); (m^n-1)\(m-1)<1,vecmax(factor(n)[,2]))}, [1..20])

a(n) = 2^A051903(n)*(m^n-1)/(m-1) with m = 2^A268336(n).

A174824 a(n) = period of the sequence {m^m, m >= 1} modulo n.

Original entry on oeis.org

1, 2, 6, 4, 20, 6, 42, 8, 18, 20, 110, 12, 156, 42, 60, 16, 272, 18, 342, 20, 42, 110, 506, 24, 100, 156, 54, 84, 812, 60, 930, 32, 330, 272, 420, 36, 1332, 342, 156, 40, 1640, 42, 1806, 220, 180, 506, 2162, 48, 294, 100, 816, 156, 2756, 54, 220, 168, 342

Offset: 1

Franklin T. Adams-Watters, Mar 30 2010

nonn

This is a divisibility sequence: if n divides m, a(n) divides a(m).

We have the equality n = a(n) for numbers n in A124240, which is related to Carmichael's function (A002322). The largest values of a(n) occur when n is prime, in which case a(n) = n*(n-1). - T. D. Noe, Feb 21 2014

			For n=3, 1^1 == 1 (mod 3), 2^2 == 1 (mod 3), 3^3 == 0 (mod 3), etc. The sequence of residues 1, 1, 0, 1, 2, 0, 1, 1, 0, ... has period 6, so a(3) = 6. - _Michael B. Porter_, Mar 13 2018

T. D. Noe, Table of n, a(n) for n = 1..1000
José María Grau and Antonio M. Oller-Marcén, On the last digit and the last non-zero digit of n^n in base b, arXiv:1203.4066 [math.NT], 2012. (See page 3)
Index to divisibility sequences

Cf. A000312, A009262, A127699.

Cf. A002322, A124240, A268336.

Table[LCM[n, CarmichaelLambda[n]], {n, 100}] (* T. D. Noe, Feb 20 2014 *)

a(n)=local(ps);ps=factor(n)[,1]~;for(k=1,#ps,n=lcm(n,ps[k]-1));n

a(n) = lcm(n, lcm(znstar(n)[2])); \\ Michel Marcus, Mar 18 2016; corrected by Michel Marcus, Nov 13 2019

apply( {A174824(n)=lcm(lcm([p-1|p<-factor(n)[,1]]),n)}, [1..99]) \\ [...] = znstar(n)[2], but 3x faster. - M. F. Hasler, Nov 13 2019

a(n) = lcm(n, A173614(n)) = lcm(n, A002322(n)) = lcm(n, A011773(n)).
If n and m are relatively prime, a(n*m) = lcm(a(n), a(m)); a(p^k) = (p-1)*p^k for p prime and k > 0.
a(n) = n*A268336(n). - M. F. Hasler, Nov 13 2019

A329126 a(n) is the least positive number which yields a multiple of n when its decimal digits (which are necessarily 0's and 1's) are read in any base.

Original entry on oeis.org

1, 110, 101010, 111100, 100010001000100010, 1111110, 10000010000010000010000010000010000010, 11111111000, 1010101010101010100, 1100110011001100110, 100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000100000000010, 11111111111100

Offset: 1

Alon Ran, Nov 05 2019

a(n) might be called the "smallest trans-basic multiple of n."
In order to be a valid binary number, the terms may contain only 0's and 1's.
The number of 1's in a(n) is conjectured to be n; the number of 0's separating each one digit is usually A268336(n)-1 for small n. The number of trailing 0's is A051903(n).
The first 21 terms have been found and verified to be minimal via an advanced search; the 13th term (see b-file) contains 146 digits, and in general for every prime n the corresponding a(n) is conjectured to contain 2+(n-1)^2 0's and 1's.
A lower bound for a(n) is given by a(A032742(n)). Proof: If a(n) were smaller than a(A032742(n)), then a(A032742(n)) would not be the smallest trans-basic multiple of A032742(n); a(n) would be. By definition a(n) is the smallest trans-basic multiple of n, so we have a contradiction; QED.
To verify a trans-basic multiple of n for n > 2, one must only: A) make sure the string has some multiple of n of '1' digits; B) make sure the string ends with at least one '0' digit; and C) check that, for all prime bases below n, the resulting number is divisible by n. If these three conditions are met, the string is a trans-basic multiple of n.
While the formula given below is guaranteed to provide a trans-basic multiple of n, it does not always yield a(n) which by definition is the smallest such number. [Corrected by M. F. Hasler, Nov 14 2019]
From N. J. A. Sloane, Nov 12 2019: (Start)
For each n, the values of (string a(n) read in base b)/n for b = 1,2,3,... give a sequence of integers.
For n=1 this is the all-1's sequence A000012.
For n=2, a(2) = 110 which in base b is b+b^2. Divided by 2 we get (b+b^2)/2, which evaluated at b = 1,2,3,4,... is 1,3,6,10,..., the triangular numbers A000217.
For n=3, we get (b+b^3+b^5)/3, which is A220892.
For n=4, we get A328994. (End)
See A329000 = (1, 6, 42, 60, 139810, 126, ...) for a(n) converted from base 2 to base 10, i.e., the numbers which yield the terms here when written in base 2. - M. F. Hasler, Nov 09 2021

			a(3) = 101010:
  101010_2  =     42 =    14*3;
  101010_3  =    273 =    91*3;
  101010_4  =   1092 =   364*3;
  101010_5  =   3255 =  1085*3;
  101010_6  =   7998 =  2666*3;
  101010_7  =  17157 =  5719*3;
  101010_8  =  33288 = 11096*3;
  101010_9  =  59787 = 19929*3;
  101010_10 = 101010 = 33670*3;
  101010_11 = 162393 = 54131*3;
  101010_12 = 250572 = 83524*3;
and so on. All the resulting values are multiples of 3.

Alon Ran, Table of n, a(n) for n = 1..21
Alon Ran, Comments on this sequence

Cf. A000012, A000217, A051903, A220892, A268336, A329338 (an upper bound), A328994.
A329000 gives a(n) read in base 2 and converted to base 10.
See also A329443.

\\ See A329338 for an upper bound which equals a(n) in many cases, e.g., all n < 14 except for n = 10. - M. F. Hasler, Nov 10 2021

To generate an upper bound on a(n), start with n 1's (this is required to ensure that it is divisible by n in bases n+1, 2n+1, etc.)
Next, place A268336(n)-1 0's in between the 1's (this ensures that the powers that are added will always sum to 0 (mod n)).
Finally, add A051903(n) 0's on the right (this is to ensure that the number will be divisible by n in bases that are roots of factors of n).
Note that this formula does not always yield the minimal solution a(n). For instance, a(10) is obtained from the above result by grouping the 1's in pairs and separating the pairs by two 0's.
a(n) <= A329338(n), with equality except for n = 10, 14, 15, ... - M. F. Hasler, Nov 14 2019

I have weakened some of the assertions in the Comments section, since they seemed to be unproved. See Alon Ran's comments (see Links). - N. J. A. Sloane, Dec 02 2019
Definition corrected, following a remark by Don Reble, by M. F. Hasler, Nov 09 2021
The present definition has been reworded by Peter Munn, Nov 17 2021, and by N. J. A. Sloane, Nov 29 2021

A329000 a(n) is the least positive number which yields a multiple of n when its binary digit string, S(n), is read in any numeric base; a(n) is displayed in base 10.

Original entry on oeis.org

1, 6, 42, 60, 139810, 126, 139620524162, 2040, 349524, 419430, 2537779500750160131246576896002, 16380, 44612382091907903486070965589630128805126146, 418861572486, 146602109610, 1048560

Offset: 1

N. J. A. Sloane, Nov 12 2019

If we consider sequence terms to be character strings, a(n) is A329126(n) read in base 2 and converted to base 10. Based on b-file for A329126.

Least k >= 1 such that n divides A329443(k). - Peter Munn, Dec 02 2021

			The strings S(1), S(2), S(3), ... are 1, 110, 101010, 111100, 100010001000100010, 1111110, ... (A329126); converted from binary to decimal these give the current sequence.

Cf. A329126, A329339, A329338, A067029, A268336, A329443.

\\ See A329339. - M. F. Hasler, Nov 14 2019

a(n) <= A329339(n) = 2^A051903(n)*(m^n-1)/(m-1) with m = 2^A268336(n), equality except for n = 10, 14, 15, ... - M. F. Hasler, Nov 14 2019

Definition corrected: not lexicographically earliest string, but smallest binary number. - M. F. Hasler, Nov 09 2021

Name aligned with new A329126 name by Peter Munn, Dec 02 2021

cp's OEIS Frontend

A267817 Numbers m that are divisible by A268336(m).

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A329338 a(n) = {1{0}^(A268336(n)-1)}^(n-1){1}{0}^A051903(n): upper bound for A329126(n).

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A329339 a(n) = 2^A051903(n)*Sum_{k=0..n-1} 2^(A268336(n)*k): upper bound for A329000(n).

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A174824 a(n) = period of the sequence {m^m, m >= 1} modulo n.

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A329126 a(n) is the least positive number which yields a multiple of n when its decimal digits (which are necessarily 0's and 1's) are read in any base.

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A329000 a(n) is the least positive number which yields a multiple of n when its binary digit string, S(n), is read in any numeric base; a(n) is displayed in base 10.

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A329339 a(n) = 2^A051903(n)Sum_{k=0..n-1} 2^(A268336(n)k): upper bound for A329000(n).